Irreps of SO(3)

The Lie algebra $𝔰 𝔬 (3)$ has $2 𝑗 + 1$ dimensional irreps $𝑑 Γ 𝑗$ for $𝑗 \in 12 ℕ 0$ , while the group $S O (3)$ only has irreps for $𝑗 \in ℕ 0$ . #m/thm/rep/lie

Proof

Let $𝐽 1, 𝐽 2, 𝐽 3 \in 𝔰 𝔬 (3)$ be the basis defined in Lie algebra, and $𝐽 2 = ⃗ 𝐽 \cdot ⃗ 𝐽$ be the quadratic Casimir element. Now consider a representation $𝑑 Γ : 𝔰 𝔬 (3) \to 𝔤 𝔩 (𝑉)$ on a finite-dimensional vector space $𝑉$ , which we will invoke implicitly. Let $| 𝑗, 𝑚 ⟩ \in 𝑉$ such that

𝐽 2 | 𝑗, 𝑚 ⟩ = 𝑗 (𝑗 + 1) | 𝑗, 𝑚 ⟩ 𝐽 3 | 𝑗, 𝑚 ⟩ = 𝑚 | 𝑗, 𝑚 ⟩

Then

𝐽 \pm = 𝐽 1 \pm 𝑖 𝐽 2

are Ladder operators of $𝐽 3$ . It follows that $𝐽 \pm | 𝑗, 𝑚 ⟩ = (𝑚 \pm 1) | 𝑗, 𝑚 \pm 1 ⟩$ transforms in the same irrep as $| 𝑗, 𝑚 ⟩$ . Since $𝑉$ is finite dimensional this must terminate at both ends, hence there exist $𝑝 < 𝑞$ such that

𝐽 3 | 𝑗, 𝑝 ⟩ = 𝑝 | 𝑗, 𝑝 ⟩ 𝐽 - | 𝑗, 𝑝 ⟩ = 0 𝐽 3 | 𝑗, 𝑞 ⟩ = 𝑞 | 𝑗, 𝑞 ⟩ 𝐽 + | 𝑗, 𝑞 ⟩ = 0

In addition since

𝐽 \mp 𝐽 \pm = (𝐽 1 \mp 𝑖 𝐽 2) (𝐽 1 \pm 𝑖 𝐽 2) = 𝐽 21 + 𝐽 22 \pm 𝑖 [𝐽 1, 𝐽 2] = 𝐽 21 + 𝐽 22 \mp 𝐽 3

it follows

𝐽 2 = 𝐽 23 + 𝐽 \mp 𝐽 \pm \pm 𝐽 3

and thus

𝐽 2 | 𝑗, 𝑝 ⟩ = (𝐽 23 - 𝐽 3 + 𝐽 + 𝐽 -) | 𝑗, 𝑝 ⟩ = 𝑝 (𝑝 - 1) | 𝑗, 𝑝 ⟩ 𝐽 2 | 𝑗, 𝑞 ⟩ = (𝐽 23 + 𝐽 3 + 𝐽 - 𝐽 +) | 𝑗, 𝑞 ⟩ = 𝑝 (𝑝 + 1) | 𝑗, 𝑞 ⟩

hence

𝑗 (𝑗 + 1) = 𝑝 (𝑝 - 1) = 𝑞 (𝑞 + 1)

and since $𝑝 < 𝑞$ we have $𝑝 = - 𝑗$ and $𝑞 = 𝑗$ . Now since $2 𝑗 = 𝑞 - 𝑝 \in ℕ 0$ , we have $2 𝑗 + 1$ dimensional irreps $𝑑 Γ 𝑗$ of $𝔰 𝔬 (3)$ labelled by $𝑗 = 0, 12, 1, \dots$ . Now assume $𝑑 Γ$ has a corresponding group representation $Γ 𝑗$ . Then

𝛿 𝑚 𝑚' ⟨ 𝑗, 𝑚 | 𝑒 - 2 𝜋 𝑖 𝐽 3 | 𝑗, 𝑚' ⟩ = ⟨ 𝑗, 𝑚 | 𝑒 - 2 𝜋 𝑖 𝑚' | 𝑗, 𝑚' ⟩ = 𝑒 - 2 𝜋 𝑖 𝑚' 𝛿 𝑚 𝑚'

which is a contradiction unless $𝑚' \in ℤ$ and thus $𝑗 \in ℕ 0$ .¹

#state/tidy | #lang/en | #SemBr

2023. Groups and representations, §6.8, pp. 93–96 ↩

Irreps of SO(3)

Footnotes